`int root(3)(x) ( "ln" x)^(2)dx`.

Answer

To solve the integral \( I = \int x^{\frac{1}{3}} (\ln x)^2 \, dx \), we will use the integration by parts method.

### Step-by-Step Solution:

1. **Identify \( u \) and \( dv \)**:
We will choose:
\[ u = (\ln x)^2 \quad \text{and} \quad dv = x^{\frac{1}{3}} \, dx \]

2. **Differentiate \( u \) and Integrate \( dv \)**:
Now, we differentiate \( u \) and integrate \( dv \):
\[ du = 2 \ln x \cdot \frac{1}{x} \, dx = \frac{2 \ln x}{x} \, dx \]
\[ v = \int x^{\frac{1}{3}} \, dx = \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4} x^{\frac{4}{3}} \]

3. **Apply Integration by Parts Formula**:
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ I = \left( \frac{3}{4} x^{\frac{4}{3}} (\ln x)^2 \right) - \int \left( \frac{3}{4} x^{\frac{4}{3}} \cdot \frac{2 \ln x}{x} \right) \, dx \]
Simplifying the integral:
\[ I = \frac{3}{4} x^{\frac{4}{3}} (\ln x)^2 - \frac{3}{2} \int x^{\frac{1}{3}} \ln x \, dx \]

4. **Evaluate the Remaining Integral**:
Let \( I_1 = \int x^{\frac{1}{3}} \ln x \, dx \). We will again use integration by parts:
- Choose \( u = \ln x \) and \( dv = x^{\frac{1}{3}} \, dx \).
- Then, \( du = \frac{1}{x} \, dx \) and \( v = \frac{3}{4} x^{\frac{4}{3}} \).

Applying integration by parts:
\[ I_1 = \left( \frac{3}{4} x^{\frac{4}{3}} \ln x \right) - \int \left( \frac{3}{4} x^{\frac{4}{3}} \cdot \frac{1}{x} \right) \, dx \]
\[ I_1 = \frac{3}{4} x^{\frac{4}{3}} \ln x - \frac{3}{4} \int x^{\frac{1}{3}} \, dx \]
\[ I_1 = \frac{3}{4} x^{\frac{4}{3}} \ln x - \frac{3}{4} \cdot \frac{3}{4} x^{\frac{4}{3}} + C \]
\[ I_1 = \frac{3}{4} x^{\frac{4}{3}} \ln x - \frac{9}{16} x^{\frac{4}{3}} + C \]

5. **Substitute Back to Find \( I \)**:
Now substituting \( I_1 \) back into the equation for \( I \):
\[ I = \frac{3}{4} x^{\frac{4}{3}} (\ln x)^2 - \frac{3}{2} \left( \frac{3}{4} x^{\frac{4}{3}} \ln x - \frac{9}{16} x^{\frac{4}{3}} \right) \]
\[ I = \frac{3}{4} x^{\frac{4}{3}} (\ln x)^2 - \frac{9}{8} x^{\frac{4}{3}} \ln x + \frac{27}{32} x^{\frac{4}{3}} + C \]

6. **Final Simplification**:
Factor out \( x^{\frac{4}{3}} \):
\[ I = x^{\frac{4}{3}} \left( \frac{3}{4} (\ln x)^2 - \frac{9}{8} \ln x + \frac{27}{32} \right) + C \]

### Final Answer:
\[ I = x^{\frac{4}{3}} \left( \frac{3}{4} (\ln x)^2 - \frac{9}{8} \ln x + \frac{27}{32} \right) + C \]